The number of non-isomorphic groups of order $p^2$, where $p$ is a prime number is:
1. 1
2. $p$
3. 2
4. $p^2$
What is simplest method to find number of non-isomorphic group? I read from various books but I don't understand the method.
I know that a finitely generated abelian group is isomorphic to direct product of cyclic group.
Best Answer
First, note any group of order $p^2$ is abelian. If $|Z(G)|=p^2$, this is automatic. If $|Z(G)|=p$, then $G/Z(G)$ has order $p$, hence is cyclic, so $G$ is abelian (a well known fact), hence actually $|Z(G)|=p^2$. So necessarily we must be in the first case. Analyzing these two cases suffices since $p$-groups have nontrivial center.
So any group of order $p^2$ is a finitely generated abelian group, hence is a direct product of cyclic subgroups by the structure theorem. The only possibilities are $\mathbb{Z}/(p^2)$ or $\mathbb{Z}/(p)\times\mathbb{Z}/(p)$, so there are only two groups of order $p^2$, up to isomorphism.